Advertisements
Advertisements
प्रश्न
Solve system of linear equations, using matrix method.
4x – 3y = 3
3x – 5y = 7
उत्तर
`[(4,-3),(3,-5)] [(x),(y)] = [(3),(7)] AX = B`
A = `[(4,-3),(3,-5)]`
`X = [(x),(y)] and B = [(3),(7)]`
Now, `abs A = [(4,-3),(3,-5)] = - 20 + 9 = - 11 ne 0`
`=> A^-1` exists and hence the given equation has a unique solution.
`therefore Adj A = [(-5,-3),(3,4)]^T = [(-5,3),(-3,4)]`
and `A^-1 = 1/abs A (Adj A)`
`= 1/-11 [(-5,3),(-3,4)]`
`X = A^-1 B`
`=> [(x),(y)] = 1/11 [(-5,-3),(3,4)] [(3),(7)]`
` = [(-6/11),(-19/11)]`
So, x = `-6/11, y = -19/11`
APPEARS IN
संबंधित प्रश्न
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
Find the value of x, if
\[\begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}0 & x & y \\ - x & 0 & z \\ - y & - z & 0\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab\end{vmatrix}\]
Prove that
\[\begin{vmatrix}\frac{a^2 + b^2}{c} & c & c \\ a & \frac{b^2 + c^2}{a} & a \\ b & b & \frac{c^2 + a^2}{b}\end{vmatrix} = 4abc\]
Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]
Prove the following identity:
\[\begin{vmatrix}2y & y - z - x & 2y \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} = \left( x + y + z \right)^3\]
Solve the following determinant equation:
Find the area of the triangle with vertice at the point:
(3, 8), (−4, 2) and (5, −1)
Find the area of the triangle with vertice at the point:
(2, 7), (1, 1) and (10, 8)
Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).
Prove that :
Prove that :
Prove that :
Prove that :
6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8
3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.
Solve each of the following system of homogeneous linear equations.
3x + y + z = 0
x − 4y + 3z = 0
2x + 5y − 2z = 0
Write the value of the determinant
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]
Evaluate \[\begin{vmatrix}4785 & 4787 \\ 4789 & 4791\end{vmatrix}\]
If |A| = 2, where A is 2 × 2 matrix, find |adj A|.
For what value of x is the matrix \[\begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}\] singular?
If \[\begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}\] , find the value of x.
If a > 0 and discriminant of ax2 + 2bx + c is negative, then
\[∆ = \begin{vmatrix}a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0\end{vmatrix} is\]
If \[\begin{vmatrix}a & p & x \\ b & q & y \\ c & r & z\end{vmatrix} = 16\] , then the value of \[\begin{vmatrix}p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r\end{vmatrix}\] is
Solve the following system of equations by matrix method:
2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6
Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Show that the following systems of linear equations is consistent and also find their solutions:
6x + 4y = 2
9x + 6y = 3
A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹1,600. School B wants to spend ₹2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
3x − y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if
Solve the following system of equations by using inversion method
x + y = 1, y + z = `5/3`, z + x = `4/3`
`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`
The value of λ, such that the following system of equations has no solution, is
`2x - y - 2z = - 5`
`x - 2y + z = 2`
`x + y + lambdaz = 3`
Let the system of linear equations x + y + az = 2; 3x + y + z = 4; x + 2z = 1 have a unique solution (x*, y*, z*). If (α, x*), (y*, α) and (x*, –y*) are collinear points, then the sum of absolute values of all possible values of α is ______.
